找出相加等於target的兩數。
解題思路
題目自帶官方解法。
Code
Java
Runtime 2 ms, Memory 42.7 MB
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class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
return null;
}
}
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Python3
Runtime 56 ms,Memory 15.2 MB
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class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
hashmap = {}
for i in range(len(nums)):
complement = target - nums[i]
if complement in hashmap:
return [i, hashmap[complement]]
hashmap[nums[i]] = i
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C#
Runtime 146 ms, Memory 44.8 MB
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public class Solution {
public int[] TwoSum(int[] nums, int target) {
Dictionary<int, int> map = new Dictionary<int, int>();
for(int i = 0; i < nums.Length; i++){
int num = nums[i];
int complement = target - num;
if (map.ContainsKey(complement)){
return new int[] {map[complement], i};
}else if(!map.ContainsKey(num)){
map.Add(num, i);
}
}
return null;
}
}
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Golang
Runtime 9 ms, Memory 4.2 MB
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func twoSum(nums []int, target int) []int {
mapNum := make(map[int]int)
for i, num := range nums{
var complement int = target - num
// check val is in map
if val, ok := mapNum[complement]; ok {
return []int{i, val}
}
mapNum[num] = i
}
return nil
}
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C++
Runtime 9 ms, Memory 11.2 MB
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class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
map<int, int> mapNum;
for(int i = 0; i < nums.size(); i++){
int num = nums[i];
int complement = target - num;
if (mapNum.count(complement) > 0){
return {mapNum[complement], i};
} else {
mapNum.insert({num, i});
}
}
return {-1, -1};
}
};
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C
Runtime ms, Memory MB
Reference
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struct number_hash {
int value;
int index;
UT_hash_handle hh;
};
void destroy_table(struct number_hash** table) {
struct number_hash* curr;
struct number_hash* tmp;
HASH_ITER(hh, *table, curr, tmp) {
HASH_DEL(*table, curr);
free(curr);
}
}
int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
struct number_hash* table = NULL;
struct number_hash* element;
int* ret = (int*) malloc(2 * sizeof(int));
int remaining;
for (int i = 0; i < numsSize; ++i) {
remaining = target - nums[i];
// Find if there has already been an element such that the sum is target
HASH_FIND_INT(table, &remaining, element);
if (element) {
ret[0] = element->index;
ret[1] = i;
break;
}
// Add the new number to the hash table if it doesn't exist already
HASH_FIND_INT(table, &nums[i], element);
if (!element) {
element = (struct number_hash *) malloc(sizeof(*element));
element->value = nums[i];
element->index = i;
HASH_ADD_INT(table, value, element);
}
}
destroy_table(&table);
*returnSize = 2;
return ret;
}
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